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Viefleur [7K]
3 years ago
5

If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the

system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K
Engineering
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer:

0.25 J/K

Explanation:

Given data in given question

heat (Q) = 100 J

temperature (T) = 400 K

to find out

the change in entropy of the given system

Solution

we use the entropy change equation here i.e  

ΔS = ΔQ / T           ...................a

Now we put the value of heat (Q) and Temperature (T) in equation a

ΔS is the entropy change, Q is heat and T is the temperature,  

so that

ΔS = 100/400 J/K

ΔS = 0.25 J/K

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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
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Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

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\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

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h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

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V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

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Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

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h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

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